衡水金卷先享题2024答案数学分科综合卷 新教材乙卷A
因为AB,D分别是四边形ABC)的外角∠MD,∠MDA的平分线,所以LADE=2∠MDA,∠DAE=号∠NAD,所以∠ADE+∠DAE=(LHDA+∠NAD)=3×240=120m.所以∠E=1809-(∠A1)+∠DA)=60P.
因为AB,D分别是四边形ABC)的外角∠MD,∠MDA的平分线,所以LADE=2∠MDA,∠DAE=号∠NAD,所以∠ADE+∠DAE=(LHDA+∠NAD)=3×240=120m.所以∠E=1809-(∠A1)+∠DA)=60P.
以f(x)的最小正周期为元.又2红=,所以u=2,所以所以fx)=2sn(2x-)f(x)=sin(2x+p),将点(行,0)的坐标代入得2×选②:因为函数f(x)的图象的一条对称轴为直线x=-吾+9...
四、解答题:本大题共6小题,共T0分。解答应写出必要的文宇说明、证明过程及演算步骤。17.(本小题满分10分)设集合A={.lx2=2x-3<0},集合B={xl2-a<x<2+a.(1)若a=2,...
22.5综合与实践测量与误差.·高2米的标杆CD的影子DE解:这种测量方法可行为2米,即CD=DE,..BD=AB.理由如下:如图,过点M作MF⊥AB于点F,设旗杆高AB为x米.过点F作!...
因此f(x)的单调递增区间是(0,),单调递减区间是(e,十o∞).…4分(2)解:由nx≤ax-,得a2--1m200≥n.2令s(x)1则有g(x)=21nx十1……………………………………5分令k(x)=-x-2lnx1,显然(...
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